package main

import "fmt"

/*
We have an array A of integers, and an array queries of queries.
For the i-th query val = queries[i][0], index = queries[i][1], we add val to A[index].  Then, the answer to the i-th query is the sum of the even values of A.
(Here, the given index = queries[i][1] is a 0-based index, and each query permanently modifies the array A.)
Return the answer to all queries.  Your answer array should have answer[i] as the answer to the i-th query.

Example 1:
Input: A = [1,2,3,4], queries = [[1,0],[-3,1],[-4,0],[2,3]]
Output: [8,6,2,4]
Explanation:
At the beginning, the array is [1,2,3,4].
After adding 1 to A[0], the array is [2,2,3,4], and the sum of even values is 2 + 2 + 4 = 8.
After adding -3 to A[1], the array is [2,-1,3,4], and the sum of even values is 2 + 4 = 6.
After adding -4 to A[0], the array is [-2,-1,3,4], and the sum of even values is -2 + 4 = 2.
After adding 2 to A[3], the array is [-2,-1,3,6], and the sum of even values is -2 + 6 = 4.

Note:
1 <= A.length <= 10000
-10000 <= A[i] <= 10000
1 <= queries.length <= 10000
-10000 <= queries[i][0] <= 10000
0 <= queries[i][1] < A.length
*/

func main() {
	fmt.Println(sumEvenAfterQueries([]int{1, 2, 3, 4}, [][]int{
		{1, 0},
		{-3, 1},
		{-4, 0},
		{2, 3},
	}))
}

func sumEvenAfterQueries(A []int, queries [][]int) []int {
	B := make([]int, 0, len(A))
	evenSum := 0
	for i := 0; i < len(A); i++ {
		if A[i]%2 == 0 {
			evenSum += A[i]
		}
	}

	for i := 0; i < len(queries); i++ {
		tmp := A[queries[i][1]]
		A[queries[i][1]] += queries[i][0]
		if tmp%2 == 0 {
			evenSum -= tmp
		}

		if A[queries[i][1]]%2 == 0 {
			evenSum += A[queries[i][1]]
		}

		B = append(B, evenSum)
	}

	return B
}
